JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Continuity and Differentiability

Let $g(x)=2x^2+5|x|-3$, so $f(x)=|g(x)|$. Since $g$ is even, work on $x\ge 0$ first: $g(x)=2x^2+5x-3=(2x-1)(x+3)$, which vanishes at $x=\frac{1}{2}$ (the root $x=-3$ is outside $x\ge 0$). By symmetry $g$ also vanishes at $x=-\frac{1}{2}$.

Continuity: $g$ is a sum of a polynomial and $5|x|$, both continuous everywhere, so $g$ — and hence $f=|g|$ — is continuous everywhere. So $m=0$.

Differentiability: for $x>0$, $g^{\prime }(x)=4x+5$; for $x<0$, $g(x)=2x^2-5x-3$ and $g^{\prime }(x)=4x-5$. At $x=0$ the right-hand derivative is $5$ and the left-hand derivative is $-5$, so $g$ itself has a corner at $x=0$. Since $g(0)=-3\ne 0$, taking the modulus does not smooth this out, so $f$ is non-differentiable at $x=0$.

At $x=\pm \frac{1}{2}$, $g$ has a simple (transversal) zero — it changes sign there rather than just touching zero — so $|g|$ has a genuine V-shaped corner at each of these points too.

So $f$ fails to be differentiable at $x=0,\frac{1}{2},-\frac{1}{2}$, giving $n=3$.

$m+n=0+3=3$.